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2

I understand how arrays decay to pointers. I understand that, for the compiler, this:

void foo(int *arg1);

is 100% equivalent to this:

void foo(int arg1[]);

Should one style be preferred over the other? I want to be consistent, but I'm having a hard time justifying either decision.

Although int main(int argc, char *argv[]) and int main(int argc, char **argv) are the same, the former seems to be much more common (correct me if I'm wrong).

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If your goal is consistency, you should use the asterisk, since that's consistent with normal (non-parameter) declarations. The other syntax should die, it causes nothing but confusion. –  Benjamin Lindley Aug 26 '12 at 4:34
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2 Answers

up vote 6 down vote accepted

I would recommend against using the [] syntax for function parameters.

The one argument in favour of using [] is that it implies, in a self-documenting manner, that the pointer is expected to point to more than one thing. For example:

void swap(int *x, int *y)
double average(int vals[], int n)

These prototypes demonstrate the semantic difference. However, the ubiquitous C string, char *, disagrees with this point (have you even seen char str[] in a function parameter list? I haven't.)

Some people like to const everything they possibly can, including pass-by-value parameters. The syntax for that when using [] (available only in C99 IIRC) is less intuitive and probably less well-known:

const char *const *const words vs. const char *const words[const]

Although I do consider that final const to be overkill, in any case.

Furthermore, the manner in which arrays decay is not completely intuitive. In particular, the fact that it is not applied recursively (char words[][] doesn't work). Especially when you start throwing in more indirection, the [] syntax just causes confusion. IMO it is better to always use pointer syntax rather than "pretending" that an array is passed as an argument.

More information: http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=aryptr#aryptrparam.

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is it more safe to use[] this syntax then * one? –  Coding Mash Aug 26 '12 at 6:40
    
What do you mean by safe? –  Mk12 Aug 26 '12 at 12:57
    
I mean safe from potential pointer hazards –  Coding Mash Aug 26 '12 at 14:19
    
There is absolutely no difference to the compiler whether you use an array or a pointer as a function parameter—it will end up being a pointer either way, and consequently be subject to "pointer hazards". –  Mk12 Aug 26 '12 at 18:45
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up vote 3 down vote

Except for char*, I use Type array[N], where N is some number or a defined constant, when the passed item conceptually is an array (i.e., it contains N>1 elements), Type * pointer when the passed item is a pointer to exactly one object.

I tend to use std::vector if the array is of a variable size. C99's concept of variable sized arrays is not available in C++.

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But the N is silently ignored, which I find misleading. –  Keith Thompson Aug 26 '12 at 7:34
    
@KeithThompson: You are correct; the compiler does ignore it. However, you aren't just writing for the compiler, you are also writing for the human user of your code. That arrays decay into pointers is one of the biggest flaws in C/C++ from the perspective of a scientific programmer. There is no reason to kowtow to that huge gaping flaw. –  David Hammen Aug 26 '12 at 12:45
    
What if you had a pointer to an array of strings (i.e. char pointer for the string, pointer to char pointer for words, pointer to pointer to char pointer to allow the function to write to it)? char *(*words[])? IMO char ***words looks better and is easier to take in at a glance. I suppose, though, that this kind of thing is more common in C (which I am primarily interested in) than C++, though. –  Mk12 Aug 26 '12 at 13:14
    
@DavidHammen: Yes, you're writing for the human user, which is why I don't like to imply that the N actually means something. –  Keith Thompson Aug 26 '12 at 18:05
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